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Not sure about this probability question I came across. Any help would be appreciated.

One bet is the event of getting at least one '1' in four rolls of a fair 6-sided die. Second bet is rolling two fair 6-sided dice 24 times with a bet on having at least one '(1,1)'. If you had to choose between playing either of these two games, which one would you choose, in order to maximize your probability of winning?

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closed as off-topic by Did, Namaste, projectilemotion, Daniel W. Farlow, Adam Hughes Mar 29 '17 at 21:18

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  • $\begingroup$ What do you know about probabilities? Have you learned about the Binomial distribution? $\endgroup$ – David Mar 6 '17 at 20:06
  • $\begingroup$ Ive done a few examples of binomial distribution when I have been given the probabilities. So here Im not given them so I'm unsure of what to do. $\endgroup$ – Shauna Mar 6 '17 at 20:09
  • $\begingroup$ You might be interested in reading about the Chevalier de Mere Problem $\endgroup$ – lulu Mar 6 '17 at 20:14
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The first scenario "at least one '1' in four rolls" can be thought of as the probability of getting at least one success in 4 attempts when a success is defined as rolling a '1'. Let us call the number of successes $N$. We first note the probability of a success on any given attempt is simply $1/6$ (as there are six outcomes that are equally likely and only one leads to a success). With this information, we have that $N$ is distributed $\text{Bin}(4,1/6)$.

Similarly, for the second scenario, we can think of it as the probability of getting one success in 24 attempts when a success is defined as rolling "snake eyes." The probability of success is then $1/36$ as there are $36$ combinations of rolls and only one of them leads to a success. Therefore in this case, $N$ is distributed $\text{Bin}(24,1/36)$. Are you able to solve from here?

EDIT:

The binomial distribution tells us the probability distribution over the number of success given you have $n$ independent trials and a probability of success of $p$. In particular, the probability of getting exactly $k$ sucesses is $${{n}\choose{k}} p^k(1-p)^{n-k}.$$

Now, in our case, we want to know the probability that we get more than zero successes. So, with a $4$ attempts to roll a $1$ we first compute the probability of getting $k=0$ successes. This is $${{4}\choose{0}} p^0(1-p)^{4-0} = 1 \cdot 1 \cdot(1-1/6)^{4} = (5/6)^4 = 0.4822.$$ Now, since probabilities add up to one, we have that the probability of getting more zero successes is simply one minus the above probability, that is $$1 - 0.4822 = 0.5178.$$ You can do the same thing for the next one and compare the answers to find out which is better.

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  • $\begingroup$ I have a formula with alot of k's and p's and I'm unsure exactly what to do. If you could show me what the formula does with one of the bets above I'm sure I can work out the other one. Thanks $\endgroup$ – Shauna Mar 6 '17 at 20:21
  • $\begingroup$ @shauna no problem, i'll add in an edit to give you an idea of how to do it. $\endgroup$ – David Mar 6 '17 at 20:24
  • $\begingroup$ I understand so much better now. Thanks very much for your help! $\endgroup$ – Shauna Mar 6 '17 at 20:36
  • $\begingroup$ @shauna no problem, glad I could be helpful! $\endgroup$ – David Mar 6 '17 at 20:37
  • $\begingroup$ Sorry I have a question about this. When you have 4 over 0, what do you do with this that equals 1? $\endgroup$ – Shauna Mar 6 '17 at 21:24
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The probability of getting at least one $1$ in four rolls is one minus the probabilty of getting no $1$'s (in 4 rolls)

\begin{eqnarray*} 1-\left(\frac{5}{6}\right)^4=0.5177 \cdots \end{eqnarray*}

The second possibility ...

\begin{eqnarray*} 1-\left(\frac{35}{36}\right)^{24}=0.4914 \cdots \end{eqnarray*}

So you are better to play the first game.

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  • $\begingroup$ Is this using the binomial distribution formula? $\endgroup$ – Shauna Mar 6 '17 at 20:22

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